/*
Count Complete Tree Nodes Total Accepted: 8843 Total Submissions: 43612 My Submissions Question Solution
Given a complete binary tree, count the number of nodes.

Definition of a complete binary tree from Wikipedia:
In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible.
It can have between 1 and 2h nodes inclusive at the last level h.

*/

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <algorithm>
#include <string>
#include <stack>
#include <queue>
#include <fstream>
#include <sstream>
#include <unordered_set>
#include <unordered_map>
#include "print.h"
using namespace std;

/**
* Definition for binary tree*/


void testForStack()
{
	stack<int> mystack;
	mystack.push(10);
	mystack.push(20);
	mystack.top() -= 5;
	cout << "mystack.top() is now " << mystack.top() << endl;
}

void testForIntToString()
{
	int a = 10;
	stringstream ss;
	ss << a;
	string str = ss.str();
	cout << str << endl;

	string str1 = to_string(a);

}


/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
	int getLeft(TreeNode *root)
	{
		int count = 0;
		while (root->left !=NULL)
		{
			root = root->left;
			++count;
		}
		return count;
	}
	
	int getRight(TreeNode* root)
	{
		int count = 0;
		while (root->right != NULL)
		{
			root = root->right;
			++count;
		}
		return count;
	
	}



	int countNodes(TreeNode* root) {
		if (root == NULL)
		{
			return 0;
		}

		int l = getLeft(root) + 1;
		int r = getRight(root)+1;

		if (l == r)
		{
			return (2 << (l - 1)) - 1;
		}
		else{
			return countNodes(root->left) + countNodes(root->right) + 1;
		}
	}
};


/**
* Your BSTIterator will be called like this:
* BSTIterator i = BSTIterator(root);
* while (i.hasNext()) cout << i.next();
*/

int main(int argc, char* argv[])
{

	char a1[] = { '1', '0', '1', '0', '0' };
	char a2[] = { '1', '0', '1', '1', '0' };
	char a3[] = { '1', '1', '1', '1', '1' };
	char a4[] = { '1', '0', '0', '1', '0' };

	vector<char> level1(a1, a1 + sizeof(a1) / sizeof(char));
	vector<char> level2(a1, a1 + sizeof(a1) / sizeof(char));
	vector<char> level3(a1, a1 + sizeof(a1) / sizeof(char));
	vector<char> level4(a1, a1 + sizeof(a1) / sizeof(char));

	vector<vector<char> > martix;
	martix.push_back(level1);
	martix.push_back(level2);
	martix.push_back(level3);
	martix.push_back(level4);

	Solution s;
	int result;
	result = s.maximalSquare(martix);


	//int n = 7;
	//ifs1.countPrimes(val);

	//strRes = s.findRepeatedDnaSequences(s1);
	//cout << "The result is : " << result << endl;
	//result = s.partition(str);
	//stackTree.push(p->left);
	//stackTree.push(p->right);
	//if (s1.containsNearbyAlmostDuplicate(vecInt, 1, 2))
	//	cout << " True" << endl;
	//else
	//cout << "false" << endl;
	system("pause");
	return 0;
}
//std::unordered_set<std::string> myset =
//{ "hot", "dot", "dog", "lot", "log" };

//std::cout << "myset contains:";
// for (auto it = myset.begin(); it != myset.end(); ++it)
//std::cout << " " << *it;
//;; std::cout << std::endl;

//TreeNode *root = new TreeNode(1);
//TreeNode *left = new TreeNode(2);
//TreeNode *right = new TreeNode(3);

//root->left = left;
//root->right = right;s
/*
//string s1 = "GAGAGAGAGAGA";
string s1 = "GAGAGAGAGAGA";
vector<string> strRes;

TreeNode *root = new TreeNode(1);
TreeNode *left1 = new TreeNode(2);
TreeNode *left2 = new TreeNode(5);

TreeNode *right1 = new TreeNode(3);
TreeNode *right2 = new TreeNode(4);

root->left = left1;
left1->right = left2;

root->right = right1;
right1->right = right2;

//vector<char> level1({ '1', '1', '1', '1', '0' });
//vector<char> level2({ '1', '1', '0', '1', '0' });
//vector<char> level3({ '1', '1', '0', '0', '0' });
//vector<char> level4({ '0', '0', '0', '0', '0' });


vector<vector<char>> grid;
grid.push_back(level1);
grid.push_back(level2);
grid.push_back(level3);
grid.push_back(level4);ss
ListNode *head = new ListNode(1);
ListNode *head1 = new ListNode(2);
ListNode *head2 = new ListNode(6);
ListNode *head3 = new ListNode(3);
ListNode *head4 = new ListNode(4);
ListNode *head5 = new ListNode(5);

ListNode *head6 = new ListNode(6);

head->next = head1;
head1->next = head2;
head2->next = head3;
head3->next = head4;
head4->next = head5;
head5->next = head6;
int val = 6;
*/
